Question 994253
Look for the negative value for a.


Distance Formula, {{{sqrt((a-2)^2+(3a-1)^2)=5}}}



{{{a^2-4a+4+9a^2-6a+1=25}}}


{{{10a^2-10a+5=25}}}


{{{10(a^2-a)=20}}}


{{{a^2-a=2}}}


{{{a^2-a-2=0}}}


{{{(a+1)(a-2)=0}}}
a=-1 will put the coordinates into quadrant #3, to be point  (-1,-3).


The other solution of a=2 will not be in quadrant #3.